diff --git a/notes/53. 数字在排序数组中出现的次数.md b/notes/53. 数字在排序数组中出现的次数.md
index f8e47b90..580507b6 100644
--- a/notes/53. 数字在排序数组中出现的次数.md	
+++ b/notes/53. 数字在排序数组中出现的次数.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/70610bf967994b22bb1c26f9ae901fa2?tpId=13&tqId=11190&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[Leetcode](https://leetcode.cn/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof/)
 
 ## 题目描述
 
@@ -41,11 +41,11 @@ public int binarySearch(int[] nums, int K) {
 但是在查找第一个位置时，找到元素之后应该继续往前找。也就是当 nums[m]\>=k 时，在左区间继续查找，左区间应该包含 m 位置。
 
 ```java
-private int binarySearch(int[] nums, int K) {
+private int binarySearch(int[] nums, int target) {
     int l = 0, h = nums.length;
     while (l < h) {
         int m = l + (h - l) / 2;
-        if (nums[m] >= K)
+        if (nums[m] >= target)
             h = m;
         else
             l = m + 1;
@@ -57,10 +57,22 @@ private int binarySearch(int[] nums, int K) {
 查找最后一个位置可以转换成寻找 k+1 的第一个位置，并再往前移动一个位置。
 
 ```java
-public int GetNumberOfK(int[] nums, int K) {
-    int first = binarySearch(nums, K);
-    int last = binarySearch(nums, K + 1);
-    return (first == nums.length || nums[first] != K) ? 0 : last - first;
+public int countTarget(int[] scores, int target) {
+    int first = binarySearch(scores, target);
+    int last = binarySearch(scores, target + 1);
+    return (first == scores.length || scores[first] != target) ? 0 : last - first;
+}
+
+private int binarySearch(int[] nums, int target) {
+    int l = 0, h = nums.length;
+    while (l < h) {
+        int m = l + (h - l) / 2;
+        if (nums[m] >= target)
+            h = m;
+        else
+            l = m + 1;
+    }
+    return l;
 }
 ```